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\begin{document}

\begin{itemize}
\item

Define the Potts action

$$- \beta \ham = \sum_{i=1}^N J \delta_{\sigma_i, \sigma_{i+1}}$$

and add the trivial field $J_0$ to obtain

$$- \beta \ham = \sum_{i=1}^N J \delta_{\sigma_i, \sigma_{i+1}} + \sum_{i=1}^N J_0$$

Now the partition function is given by

$$Z_N = \sum_{\sigma_1 = 1}^s \sum_{\sigma_2 = 1}^s \cdots\sum_{\sigma_N = 1}^s \exp \left\{ \sum_{i=1}^N J \delta_{\sigma_i, \sigma_{i+1}} + \sum_{i=1}^N J_0 \right\} = \sum_{\{\sigma\}} \prod_{i=1}^N e^{w(\sigma_i,\sigma_{i+1})}$$

Under a $\lambda = 2$ decimation every second spin is retained, and the intermediate spins summed out

$$Z_N = \sum_{\sigma_1 = 1}^s \sum_{\sigma_3 = 1}^s \cdots \sum_{\sigma_N = 1}^s \left[ \sum_{\sigma_2 = 1}^s e^{w(\sigma_1,\sigma_2)} e^{w(\sigma_2,\sigma_3)} \right] \left[ \sum_{\sigma_3 = 1}^s e^{w(\sigma_2,\sigma_3)} e^{w(\sigma_3,\sigma_4)} \right] \cdots $$
$$= \sum_{\sigma_1 = 1}^s \sum_{\sigma_3 = 1}^s \cdots \sum_{\sigma_N = 1}^s e^{w'(\sigma_1,\sigma_3)} e^{w'(\sigma_3,\sigma_4)} \cdots$$

so the partition function sum retains the same form, but for a renormalised interaction $w'$. Since this is the same for every unit of 3 spins, we need only study one representative unit

$$e^{J'_0} e^{J' \delta_{\sigma,\sigma'}} = e^{2J_0} \sum_{\mu = 1}^s e^{J \delta_{\sigma,\mu}} e^{J \delta_{\mu,\sigma'}}$$

To evaluate the sum, consider the two general cases $\sigma = \sigma'$ and $\sigma \neq \sigma'$ separately. In the first case we have

$$e^{w'(\sigma,\sigma)} = e^{J'_0} e^{J'} = e^{2J_0} \sum_{\mu = 1}^s e^{J \delta_{\sigma,\mu}} e^{J \delta_{\mu,\sigma}} = e^{2J_0} \sum_{\mu = 1}^s e^{2 J \delta_{\mu, \sigma}} = e^{2J_0} [ e^{2J} + (s - 1) ]$$

where we have one $e^{2J}$ term from the $\mu = \sigma$ term, and $(s-1)$ contributions of $e^0=1$ where $\mu \neq \sigma$. In the second case

$$e^{w'(\sigma,\sigma')} = e^{J'_0} = e^{2J_0} \sum_{\mu = 1}^s e^{J \delta_{\sigma,\mu}} e^{J \delta_{\mu,\sigma'}}= e^{2J_0} \sum_{\mu = 1}^s e^{2 J \delta_{\mu, \sigma}} = e^{2J_0} [ e^{J}e^0 + e^0 e^J + (s - 2)e^0 ] = e^{2J_0} [ 2e^J + (s - 2) ]$$

we thus have 2 equations relating our two couplings $J$, $J_0$

$$e^{J'_0} e^{J'} = e^{2J_0} [ e^{2J} + (s - 1) ]$$
$$e^{J'_0} = e^{2J_0} [ 2e^J + (s - 2) ]$$

the quotient of which gives the RG equation for $J$

$$e^{J'} = \frac{  e^{2J} + (s - 1) }{ 2e^J + (s - 2) }$$

and we disregard the equation for $J'_0$.

\item

To find the fixed points, defining $x = e^{J^\star}$ we have a quadratic equation for $x$

$$x = \frac{x^2 + (s - 1)}{2x + (s-2)} \Rightarrow x^2 + x(s - 2) + (1-s) = 0$$

with solutions $x = 1$ and $x = 1 - s$, i.e.

$$e^{J^\star} = 1 \Rightarrow J^\star = 0$$
$$e^{J^\star} = 1 - s \Rightarrow J^\star = \log(1-s)$$

where the second possibility is unphysical, since $s = 1$ corresponds to a system with only one state and thus no dynamics. There is thus only one physical fixed point at $J^\star = 0$, corresponding to the completely disordered phase.

We expect the RG flow to be towards the disordered fixed point. To check this, consider $J \to \infty$, then we can neglect other terms in the RG equation

$$e^{J'} \sim \frac{e^{2J}}{2e^J} = \frac{1}{2} e^J \Rightarrow J' = J - \log 2$$

so under rescaling $J$ is decreased; the system moves towards the disordered $J=0$ fixed point.

\item

Setting $s=2$ in the RG equation for $J$

$$e^{J'} = \frac{  e^{2J} + (2 - 1) }{ 2e^J + (2 - 2) } = \frac{1}{2} ( e^{2J} + 1 ) e^{-J} = \cosh J$$

The Ising RG equation for $\lambda = 2$ decimation is (see Q1)

$$\tanh K' = \tanh^2 K$$

The $s=2$ Potts model is very similar to the Ising model, since they each have two states per spin and a ferromagnetic coupling between nearest neighbours. There is some difference however; in the Ising model the energy cost for a neighbouring pair to be anti-aligned is $K - (-K) = 2K$, whereas in the Potts model the cost is $J - 0 = J$. Under the appropriate mapping from $J \to K$ it should be possible to recover the Ising RG equation from the Potts equation. Following the insight concerning energy differences, we set $J' = 2 K'$, $J = 2K$ in the Potts RG equation

$$e^{2K'} = \cosh(2K) \implies e^{K'} = \sqrt{ \cosh(2K) }$$

then denote $x=\sqrt{ \cosh(2K) }$ for brevity, and construct $\tanh K'$

$$2 \sinh K' = e^{K'} - e^{-K'} = x - 1/x$$
$$2 \cosh K' = e^{K'} + e^{-K'} = x + 1/x$$

$$\implies \tanh K' = \frac{ x - 1/x }{ x + 1/x } = \frac{ x^2 - 1 }{ x^2 + 1 } = \frac{\cosh(2K) - 1}{\cosh(2K) + 1} = \tanh^2 K$$

which is the Ising RG equation, and the last equality follows from the double angle formulae $2 \cosh^2 x = (\cosh(2x) + 1)$, $2 \sinh^2 x = (\cosh(2x) - 1)$.

\item



\end{itemize}
\end{document}